Question

Equation of two diameters of a circle are \(2x-3y=5\) and \(3x-4y=7\).The line joining the points \((-\frac{22}{7},-4)\) and \((-\frac{1}{7},3)\) intersects the circle at only one point \(P(\alpha,\beta)\).Then \(17\beta-\alpha\) is equal to.

Answer 1

Correct Answer: 2

Step 1: Determine the circle's center.

The intersection of the diameters yields the circle's center. We solve the following system of linear equations: \[ 2x - 3y = 5 \quad \text{and} \quad 3x - 4y = 7 \]

To eliminate \( x \), multiply the first equation by 3 and the second by 2: \[ 6x - 9y = 15 \\ 6x - 8y = 14 \] Subtract the second resulting equation from the first: \[ (6x - 9y) - (6x - 8y) = 15 - 14 \] \[ -y = 1 \implies y = -1 \]

Substitute \( y = -1 \) into the equation \( 2x - 3y = 5 \):

\[ 2x - 3(-1) = 5 \implies 2x + 3 = 5 \implies 2x = 2 \implies x = 1 \]

Therefore, the center of the circle is located at: \[ C(1, -1) \]

Step 2: Derive the equation of the line connecting the given points.

Let the two points be: \[ A\left(-\frac{22}{7}, -4\right) \quad \text{and} \quad B\left(-\frac{1}{7}, 3\right) \]

Calculate the slope of line \( AB \): \[ m = \frac{3 - (-4)}{-\frac{1}{7} - (-\frac{22}{7})} = \frac{7}{\frac{21}{7}} = \frac{7}{3} \]

Use the point-slope form with point \( A(x_1, y_1) \) to find the line's equation: \[ y + 4 = \frac{7}{3}\left(x + \frac{22}{7}\right) \]

Simplify the equation: \[ 3(y + 4) = 7\left(x + \frac{22}{7}\right) \] \[ 3y + 12 = 7x + 22 \] \[ 7x - 3y + 10 = 0 \]

Hence, the equation of the line is determined to be: \[ 7x - 3y + 10 = 0 \]

Step 3: Establish the condition for the line to be tangent to the circle.

The line \( 7x - 3y + 10 = 0 \) is tangent to the circle. This implies that the perpendicular distance from the circle's center \( (1, -1) \) to this line is equal to the circle's radius \( r \).

The formula for the distance \( d \) from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is: \[ d = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}} \]

Substitute the values \( a = 7, b = -3, c = 10, \) and \( (x_1, y_1) = (1, -1) \): \[ d = \frac{|7(1) - 3(-1) + 10|}{\sqrt{7^2 + (-3)^2}} = \frac{|7 + 3 + 10|}{\sqrt{49 + 9}} = \frac{20}{\sqrt{58}} \] \[ r = \frac{20}{\sqrt{58}} \]

Step 4: Formulate the equation of the circle.

The equation of a circle with center \( (h, k) \) and radius \( r \) is \( (x - h)^2 + (y - k)^2 = r^2 \). Using the derived center \( (1, -1) \) and radius squared \( r^2 = \left(\frac{20}{\sqrt{58}}\right)^2 = \frac{400}{58} = \frac{200}{29} \):

\[ (x - 1)^2 + (y + 1)^2 = r^2 = \frac{400}{58} = \frac{200}{29} \] \] \[ (x - 1)^2 + (y + 1)^2 = \frac{200}{29} \]

Step 5: Identify the point of tangency (P).

The equations for the circle and the tangent line are: \[ (x - 1)^2 + (y + 1)^2 = \frac{200}{29} \] \[ 7x - 3y + 10 = 0 \] From the tangent line equation, we can express \( y \) in terms of \( x \): \[ \Rightarrow y = \frac{7x + 10}{3} \]

Substitute this expression for \( y \) into the circle's equation: \[ (x - 1)^2 + \left(\frac{7x + 10}{3} + 1\right)^2 = \frac{200}{29} \] \[ (x - 1)^2 + \left(\frac{7x + 13}{3}\right)^2 = \frac{200}{29} \]

Expand and simplify the equation: \[ 9(x - 1)^2 + (7x + 13)^2 = \frac{1800}{29} \] \[ 9(x^2 - 2x + 1) + (49x^2 + 182x + 169) = \frac{1800}{29} \] \[ 58x^2 + 164x + 178 = \frac{1800}{29} \] Multiply by 29 to clear the fraction: \[ 1682x^2 + 4756x + 5162 = 1800 \] Rearrange into a quadratic equation: \[ 1682x^2 + 4756x + 3362 = 0 \] \] Since the line is tangent, the discriminant of this quadratic equation must be zero. This indicates a single point of contact \( P(\alpha, \beta) \).

To find the x-coordinate of the point of tangency, \( \alpha \), we use the formula for the x-coordinate of the vertex of a parabola (which corresponds to the repeated root of the quadratic equation): \[ \alpha = -\frac{b}{2a} = -\frac{4756}{2(1682)} = -\frac{4756}{3364} = -\frac{1189}{841} \]

Now, find the y-coordinate, \( \beta \), by substituting \( \alpha \) back into the line equation \( y = \frac{7x + 10}{3} \): \[ \beta = \frac{7\left(-\frac{1189}{841}\right) + 10}{3} = \frac{\frac{-8323}{841} + \frac{8410}{841}}{3} = \frac{\frac{87}{841}}{3} = \frac{87}{2523} = \frac{29}{841} \]

Step 6: Compute \( 17\beta - \alpha \).

Using the calculated values of \( \alpha \) and \( \beta \): \[ 17\beta - \alpha = 17\left(\frac{29}{841}\right) - \left(-\frac{1189}{841}\right) \] \[ = \frac{493}{841} + \frac{1189}{841} = \frac{1682}{841} = 2 \]

Final Answer:

The result of the expression \( 17\beta - \alpha \) is: \[ \boxed{17\beta - \alpha = 2} \]